Problem: You have found the following ages (in years) of all 5 lizards at your local zoo: $ 2,\enspace 1,\enspace 5,\enspace 1,\enspace 2$ What is the average age of the lizards at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 5 lizards at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{2 + 1 + 5 + 1 + 2}{{5}} = {2.2\text{ years old}} $ Find the squared deviations from the mean for each lizard. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $2$ years $-0.2$ years $0.04$ years $^2$ $1$ year $-1.2$ years $1.44$ years $^2$ $5$ years $2.8$ years $7.84$ years $^2$ $1$ year $-1.2$ years $1.44$ years $^2$ $2$ years $-0.2$ years $0.04$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0.04} + {1.44} + {7.84} + {1.44} + {0.04}} {{5}} $ $ {\sigma^2} = \dfrac{{10.8}}{{5}} = {2.16\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{2.16\text{ years}^2}} = {1.5\text{ years}} $ The average lizard at the zoo is 2.2 years old. There is a standard deviation of 1.5 years.